【Paper】2013_基于一致性理论的无人机编队控制器设计

原文地址： [1]郭伟强. 基于一致性理论的无人机编队控制器设计[D].哈尔滨工业大学,2013.

2013_基于一致性理论的无人机编队控制器设计_郭伟强

3.3 一致性理论
3.4 控制方案设计
- 3.4.1 一致性协议设计
- 3.4.2 稳定性分析
3.5 仿真实验与分析
- 3.5.1 队形保持

3.3 一致性理论

所谓一致性，从控制理论的角度来说，就是指各智能体的状态变量在一定的控制协议和控制器的作用下，最终达到一致。

3.4 控制方案设计

3.4.1 一致性协议设计

本章中我们设计具有时变通信时滞的控制协议为
ui=q˙d−k1∑pi∈Ni(pi(t−τ(t))−pj(t−τ(t))−rij)−k2∑qj∈Niaij(qi(t−τ(t))−qj(t−τ(t)))−k3hi(qi−qd)(3-3)\begin{aligned} u_i &= \dot{q}_d \\ &- k_1 \sum_{p_i\in N_i} (p_i (t-\tau(t)) - p_j (t-\tau(t)) - r_{ij}) \\ &- k_2 \sum_{q_j\in N_i} a_{ij} (q_i(t-\tau(t)) - q_j(t-\tau(t))) \\ &- k_3 h_i (q_i - q_d) \tag{3-3} \end{aligned}ui=q˙d−k1pi∈Ni∑(pi(t−τ(t))−pj(t−τ(t))−rij)−k2qj∈Ni∑aij(qi(t−τ(t))−qj(t−τ(t)))−k3hi(qi−qd)(3-3)

可能是由于编写时不小心，感觉协议中应该再加上红色部分 aij\red{a_{ij}}aij 才正确（2021-05-28）。修改版如下：
ui=q˙d−k1∑pi∈Niaij(pi(t−τ(t))−pj(t−τ(t))−rij)−k2∑qj∈Niaij(qi(t−τ(t))−qj(t−τ(t)))−k3hi(qi−qd)(3-3)\begin{aligned} u_i &= \dot{q}_d \\ &- k_1 \sum_{p_i\in N_i} \red{a_{ij}} (p_i (t-\tau(t)) - p_j (t-\tau(t)) - r_{ij}) \\ &- k_2 \sum_{q_j\in N_i} a_{ij} (q_i(t-\tau(t)) - q_j(t-\tau(t))) \\ &- k_3 h_i (q_i - q_d) \tag{3-3} \end{aligned}ui=q˙d−k1pi∈Ni∑aij(pi(t−τ(t))−pj(t−τ(t))−rij)−k2qj∈Ni∑aij(qi(t−τ(t))−qj(t−τ(t)))−k3hi(qi−qd)(3-3)

精简一下，暂时忽略掉时滞作用，方便观察协议的关键因素：
ui=q˙d−k1∑pi∈Ni(pi−pj−rij)−k2∑qj∈Niaij(qi−qj)−k3hi(qi−qd)(3-3)\begin{aligned} u_i &= \dot{q}_d \\ &- k_1 \sum_{p_i\in N_i} (p_i - p_j - r_{ij}) \\ &- k_2 \sum_{q_j\in N_i} a_{ij} (q_i - q_j) \\ &- k_3 h_i (q_i - q_d) \tag{3-3} \end{aligned}ui=q˙d−k1pi∈Ni∑(pi−pj−rij)−k2qj∈Ni∑aij(qi−qj)−k3hi(qi−qd)(3-3)

Suppose n=4n=4n=4，then
[u1u2u3u4]=[q˙dq˙dq˙dq˙d]−k1[a12(p1−p2−r12)+a13(p1−p3−r13)+a14(p1−p4−r14)a21(p2−p1−r21)+a23(p2−p3−r23)+a24(p2−p4−r24)a31(p3−p1−r31)+a32(p3−p2−r32)+a34(p3−p4−r34)a41(p4−p1−r41)+a42(p4−p2−r42)+a43(p4−p3−r43)]−k2[a12(q1−q2)+a13(q1−q3)+a14(q1−q4)a21(q2−q1)+a23(q2−q3)+a24(q2−q4)a31(q3−q1)+a32(q3−q2)+a34(q3−q4)a41(q4−q1)+a42(q4−q2)+a43(q4−q3)]−k3[h1(q1−qd)h2(q2−qd)h3(q3−qd)h4(q4−qd)]\begin{aligned} \left[\begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ \end{matrix}\right] &= \left[\begin{matrix} \dot{q}_d \\ \dot{q}_d \\ \dot{q}_d \\ \dot{q}_d \\ \end{matrix}\right] \\ &-k_1 \left[\begin{matrix} a_{12}(p_1 - p_2 - r_{12}) + a_{13}(p_1 - p_3 - r_{13}) + a_{14}(p_1 - p_4 - r_{14}) \\ a_{21}(p_2 - p_1 - r_{21}) + a_{23}(p_2 - p_3 - r_{23}) + a_{24}(p_2 - p_4 - r_{24}) \\ a_{31}(p_3 - p_1 - r_{31}) + a_{32}(p_3 - p_2 - r_{32}) + a_{34}(p_3 - p_4 - r_{34}) \\ a_{41}(p_4 - p_1 - r_{41}) + a_{42}(p_4 - p_2 - r_{42}) + a_{43}(p_4 - p_3 - r_{43}) \\ \end{matrix}\right] \\ &-k_2 \left[\begin{matrix} a_{12}(q_1 - q_2) + a_{13}(q_1 - q_3) + a_{14}(q_1 - q_4) \\ a_{21}(q_2 - q_1) + a_{23}(q_2 - q_3) + a_{24}(q_2 - q_4) \\ a_{31}(q_3 - q_1) + a_{32}(q_3 - q_2) + a_{34}(q_3 - q_4) \\ a_{41}(q_4 - q_1) + a_{42}(q_4 - q_2) + a_{43}(q_4 - q_3) \\ \end{matrix}\right] \\ &-k_3 \left[\begin{matrix} h_1(q_1 - q_d) \\ h_2(q_2 - q_d) \\ h_3(q_3 - q_d) \\ h_4(q_4 - q_d) \\ \end{matrix}\right] \\ \end{aligned}⎣⎢⎢⎡u1u2u3u4⎦⎥⎥⎤=⎣⎢⎢⎡q˙dq˙dq˙dq˙d⎦⎥⎥⎤−k1⎣⎢⎢⎡a12(p1−p2−r12)+a13(p1−p3−r13)+a14(p1−p4−r14)a21(p2−p1−r21)+a23(p2−p3−r23)+a24(p2−p4−r24)a31(p3−p1−r31)+a32(p3−p2−r32)+a34(p3−p4−r34)a41(p4−p1−r41)+a42(p4−p2−r42)+a43(p4−p3−r43)⎦⎥⎥⎤−k2⎣⎢⎢⎡a12(q1−q2)+a13(q1−q3)+a14(q1−q4)a21(q2−q1)+a23(q2−q3)+a24(q2−q4)a31(q3−q1)+a32(q3−q2)+a34(q3−q4)a41(q4−q1)+a42(q4−q2)+a43(q4−q3)⎦⎥⎥⎤−k3⎣⎢⎢⎡h1(q1−qd)h2(q2−qd)h3(q3−qd)h4(q4−qd)⎦⎥⎥⎤

其中，第二部分
[a12(p1−p2−r12)+a13(p1−p3−r13)+a14(p1−p4−r14)a21(p2−p1−r21)+a23(p2−p3−r23)+a24(p2−p4−r24)a31(p3−p1−r31)+a32(p3−p2−r32)+a34(p3−p4−r34)a41(p4−p1−r41)+a42(p4−p2−r42)+a43(p4−p3−r43)]=[a12(p1−p2)−a12(r12)+a13(p1−p3)−a13(r13)+a14(p1−p4)−a14(r14)a21(p2−p1)−a21(r21)+a23(p2−p3)−a23(r23)+a24(p2−p4)−a24(r24)a31(p3−p1)−a31(r31)+a32(p3−p2)−a32(r32)+a34(p3−p4)−a34(r34)a41(p4−p1)−a41(r41)+a42(p4−p2)−a42(r42)+a43(p4−p3)−a43(r43)]=[a12(p1−p2)+a13(p1−p3)+a14(p1−p4)−a12(r12)−a13(r13)−a14(r14)a21(p2−p1)+a23(p2−p3)+a24(p2−p4)−a21(r21)−a23(r23)−a24(r24)a31(p3−p1)+a32(p3−p2)+a34(p3−p4)−a31(r31)−a32(r32)−a34(r34)a41(p4−p1)+a42(p4−p2)+a43(p4−p3)−a41(r41)−a42(r42)−a43(r43)]=[a12(p1−p2)+a13(p1−p3)+a14(p1−p4)a21(p2−p1)+a23(p2−p3)+a24(p2−p4)a31(p3−p1)+a32(p3−p2)+a34(p3−p4)a41(p4−p1)+a42(p4−p2)+a43(p4−p3)]−[a12(r12)+a13(r13)+a14(r14)a21(r21)+a23(r23)+a24(r24)a31(r31)+a32(r32)+a34(r34)a41(r41)+a42(r42)+a43(r43)]=L∗p−[a12(r12)+a13(r13)+a14(r14)a21(r21)+a23(r23)+a24(r24)a31(r31)+a32(r32)+a34(r34)a41(r41)+a42(r42)+a43(r43)]\begin{aligned} & \left[\begin{matrix} a_{12}(p_1 - p_2 - r_{12}) + a_{13}(p_1 - p_3 - r_{13}) + a_{14}(p_1 - p_4 - r_{14}) \\ a_{21}(p_2 - p_1 - r_{21}) + a_{23}(p_2 - p_3 - r_{23}) + a_{24}(p_2 - p_4 - r_{24}) \\ a_{31}(p_3 - p_1 - r_{31}) + a_{32}(p_3 - p_2 - r_{32}) + a_{34}(p_3 - p_4 - r_{34}) \\ a_{41}(p_4 - p_1 - r_{41}) + a_{42}(p_4 - p_2 - r_{42}) + a_{43}(p_4 - p_3 - r_{43}) \\ \end{matrix}\right] \\ &= \left[\begin{matrix} a_{12}(p_1 - p_2) - a_{12}(r_{12}) + a_{13}(p_1 - p_3) - a_{13}(r_{13}) + a_{14}(p_1 - p_4) -a_{14}( r_{14}) \\ a_{21}(p_2 - p_1) - a_{21}(r_{21}) + a_{23}(p_2 - p_3) - a_{23}(r_{23}) + a_{24}(p_2 - p_4) - a_{24}(r_{24}) \\ a_{31}(p_3 - p_1) - a_{31}(r_{31}) + a_{32}(p_3 - p_2) - a_{32}(r_{32}) + a_{34}(p_3 - p_4) - a_{34}(r_{34}) \\ a_{41}(p_4 - p_1) - a_{41}(r_{41}) + a_{42}(p_4 - p_2) - a_{42}(r_{42}) + a_{43}(p_4 - p_3) - a_{43}(r_{43}) \\ \end{matrix}\right] \\ &= \left[\begin{matrix} a_{12}(p_1 - p_2) + a_{13}(p_1 - p_3) + a_{14}(p_1 - p_4) ~~~ - a_{12}(r_{12})- a_{13}(r_{13}) -a_{14}( r_{14}) \\ a_{21}(p_2 - p_1) + a_{23}(p_2 - p_3) + a_{24}(p_2 - p_4) ~~~ - a_{21}(r_{21})- a_{23}(r_{23}) - a_{24}(r_{24}) \\ a_{31}(p_3 - p_1) + a_{32}(p_3 - p_2) + a_{34}(p_3 - p_4) ~~~ - a_{31}(r_{31}) - a_{32}(r_{32}) - a_{34}(r_{34}) \\ a_{41}(p_4 - p_1) + a_{42}(p_4 - p_2) + a_{43}(p_4 - p_3) ~~~ - a_{41}(r_{41}) - a_{42}(r_{42}) - a_{43}(r_{43}) \\ \end{matrix}\right] \\ &= \left[\begin{matrix} a_{12}(p_1 - p_2) + a_{13}(p_1 - p_3) + a_{14}(p_1 - p_4) \\ a_{21}(p_2 - p_1) + a_{23}(p_2 - p_3) + a_{24}(p_2 - p_4) \\ a_{31}(p_3 - p_1) + a_{32}(p_3 - p_2) + a_{34}(p_3 - p_4) \\ a_{41}(p_4 - p_1) + a_{42}(p_4 - p_2) + a_{43}(p_4 - p_3) \\ \end{matrix}\right] - \left[\begin{matrix} a_{12}(r_{12}) + a_{13}(r_{13}) + a_{14}( r_{14}) \\ a_{21}(r_{21}) + a_{23}(r_{23}) + a_{24}(r_{24}) \\ a_{31}(r_{31}) + a_{32}(r_{32}) + a_{34}(r_{34}) \\ a_{41}(r_{41}) + a_{42}(r_{42}) + a_{43}(r_{43}) \\ \end{matrix}\right] \\ &= L*p- \left[\begin{matrix} a_{12}(r_{12}) + a_{13}(r_{13}) + a_{14}( r_{14}) \\ a_{21}(r_{21}) + a_{23}(r_{23}) + a_{24}(r_{24}) \\ a_{31}(r_{31}) + a_{32}(r_{32}) + a_{34}(r_{34}) \\ a_{41}(r_{41}) + a_{42}(r_{42}) + a_{43}(r_{43}) \\ \end{matrix}\right] \\ \end{aligned}⎣⎢⎢⎡a12(p1−p2−r12)+a13(p1−p3−r13)+a14(p1−p4−r14)a21(p2−p1−r21)+a23(p2−p3−r23)+a24(p2−p4−r24)a31(p3−p1−r31)+a32(p3−p2−r32)+a34(p3−p4−r34)a41(p4−p1−r41)+a42(p4−p2−r42)+a43(p4−p3−r43)⎦⎥⎥⎤=⎣⎢⎢⎡a12(p1−p2)−a12(r12)+a13(p1−p3)−a13(r13)+a14(p1−p4)−a14(r14)a21(p2−p1)−a21(r21)+a23(p2−p3)−a23(r23)+a24(p2−p4)−a24(r24)a31(p3−p1)−a31(r31)+a32(p3−p2)−a32(r32)+a34(p3−p4)−a34(r34)a41(p4−p1)−a41(r41)+a42(p4−p2)−a42(r42)+a43(p4−p3)−a43(r43)⎦⎥⎥⎤=⎣⎢⎢⎡a12(p1−p2)+a13(p1−p3)+a14(p1−p4) −a12(r12)−a13(r13)−a14(r14)a21(p2−p1)+a23(p2−p3)+a24(p2−p4) −a21(r21)−a23(r23)−a24(r24)a31(p3−p1)+a32(p3−p2)+a34(p3−p4) −a31(r31)−a32(r32)−a34(r34)a41(p4−p1)+a42(p4−p2)+a43(p4−p3) −a41(r41)−a42(r42)−a43(r43)⎦⎥⎥⎤=⎣⎢⎢⎡a12(p1−p2)+a13(p1−p3)+a14(p1−p4)a21(p2−p1)+a23(p2−p3)+a24(p2−p4)a31(p3−p1)+a32(p3−p2)+a34(p3−p4)a41(p4−p1)+a42(p4−p2)+a43(p4−p3)⎦⎥⎥⎤−⎣⎢⎢⎡a12(r12)+a13(r13)+a14(r14)a21(r21)+a23(r23)+a24(r24)a31(r31)+a32(r32)+a34(r34)a41(r41)+a42(r42)+a43(r43)⎦⎥⎥⎤=L∗p−⎣⎢⎢⎡a12(r12)+a13(r13)+a14(r14)a21(r21)+a23(r23)+a24(r24)a31(r31)+a32(r32)+a34(r34)a41(r41)+a42(r42)+a43(r43)⎦⎥⎥⎤

A∗R=[0a12a13a14a210a23a24a31a320a34a41a42a430][0r12r13r14r210r23r24r31r320r34r41r42r430]=[a12r21+a13r31+a14r41a13r32+a14r42a12r23+a14r43a12r24+a13r34a23r31+a24r41a21r12+a23r32+a24r42a21r23+a24r43a21r14+a23r34⋯⋯]\begin{aligned} A*R &= \left[\begin{matrix} 0 & a_{12} & a_{13} & a_{14} \\ a_{21} & 0 & a_{23} & a_{24} \\ a_{31} & a_{32} & 0 & a_{34} \\ a_{41} & a_{42} & a_{43} & 0 \\ \end{matrix}\right] \left[\begin{matrix} 0 & r_{12} & r_{13} & r_{14} \\ r_{21} & 0 & r_{23} & r_{24} \\ r_{31} & r_{32} & 0 & r_{34} \\ r_{41} & r_{42} & r_{43} & 0 \\ \end{matrix}\right] \\ &= \left[\begin{matrix} a_{12}r_{21} + a_{13}r_{31} + a_{14}r_{41} & a_{13}r_{32} + a_{14}r_{42} & a_{12}r_{23} + a_{14}r_{43} & a_{12}r_{24} + a_{13}r_{34} \\ a_{23}r_{31} + a_{24}r_{41} & a_{21}r_{12} + a_{23}r_{32} + a_{24}r_{42} & a_{21}r_{23} + a_{24}r_{43} & a_{21}r_{14} + a_{23}r_{34} \\ \cdots \\ \cdots \\ \end{matrix}\right] \\ \end{aligned}A∗R=⎣⎢⎢⎡0a21a31a41a120a32a42a13a230a43a14a24a340⎦⎥⎥⎤⎣⎢⎢⎡0r21r31r41r120r32r42r13r230r43r14r24r340⎦⎥⎥⎤=⎣⎢⎢⎡a12r21+a13r31+a14r41a23r31+a24r41⋯⋯a13r32+a14r42a21r12+a23r32+a24r42a12r23+a14r43a21r23+a24r43a12r24+a13r34a21r14+a23r34⎦⎥⎥⎤

diag(A∗R)=[a12r21+a13r31+a14r41⋯⋯⋯⋯a21r12+a23r32+a24r42⋯⋯⋯⋯]\begin{aligned} &\text{diag}{(A*R)} \\ &= \left[\begin{matrix} a_{12}r_{21} + a_{13}r_{31} + a_{14}r_{41} & \cdots & \cdots & \cdots \\ \cdots & a_{21}r_{12} + a_{23}r_{32} + a_{24}r_{42} & \cdots & \cdots \\ \cdots \\ \cdots \\ \end{matrix}\right] \\ \end{aligned}diag(A∗R)=⎣⎢⎢⎡a12r21+a13r31+a14r41⋯⋯⋯⋯a21r12+a23r32+a24r42⋯⋯⋯⋯⎦⎥⎥⎤

这里，diag(⋅)\text{diag}(\cdot)diag(⋅) 表示取对角元素的意思。并且，若想使得 diag(AR)\text{diag}(AR)diag(AR) 部分成立，还默认假定了 RRR 中元素 r12r_{12}r12 与 r21r_{21}r21 相等，即 rij=rjir_{ij} = r_{ji}rij=rji。

第四部分
[h1(q1−qd)h2(q2−qd)h3(q3−qd)h4(q4−qd)]=[h1h2h3h4][q1−qdq2−qdq3−qdq4−qd]=[h1h2h3h4]([q1q2q3q4]−qd⊗[1111])=H∗(q−qd⊗1n)\begin{aligned} &\left[\begin{matrix} h_1(q_1 - q_d) \\ h_2(q_2 - q_d) \\ h_3(q_3 - q_d) \\ h_4(q_4 - q_d) \\ \end{matrix}\right] \\ &= \left[\begin{matrix} h_1 & & & \\ & h_2 & & \\ & & h_3 & \\ & & & h_4 \\ \end{matrix}\right] \left[\begin{matrix} q_1 - q_d \\ q_2 - q_d \\ q_3 - q_d \\ q_4 - q_d \\ \end{matrix}\right] \\ &= \left[\begin{matrix} h_1 & & & \\ & h_2 & & \\ & & h_3 & \\ & & & h_4 \\ \end{matrix}\right] (\left[\begin{matrix} q_1 \\ q_2 \\ q_3 \\ q_4 \\ \end{matrix}\right]- q_d \otimes \left[\begin{matrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{matrix}\right]) \\ &= H * (q - q_d \otimes 1_n) \end{aligned}⎣⎢⎢⎡h1(q1−qd)h2(q2−qd)h3(q3−qd)h4(q4−qd)⎦⎥⎥⎤=⎣⎢⎢⎡h1h2h3h4⎦⎥⎥⎤⎣⎢⎢⎡q1−qdq2−qdq3−qdq4−qd⎦⎥⎥⎤=⎣⎢⎢⎡h1h2h3h4⎦⎥⎥⎤(⎣⎢⎢⎡q1q2q3q4⎦⎥⎥⎤−qd⊗⎣⎢⎢⎡1111⎦⎥⎥⎤)=H∗(q−qd⊗1n)

改写成向量和矩阵的形式：
u=q˙d⊗1n−k1(Lp−diag(AR))−k2Lq−k3H(qi−qd⊗1n)(3-4)\begin{aligned} u &= \dot{q}_d \otimes 1_n \\ &-k_1 (Lp - \text{diag}(AR)) \\ &-k_2 Lq \\ &-k_3 H (q_i - q_d \otimes 1_n) \end{aligned} \tag{3-4}u=q˙d⊗1n−k1(Lp−diag(AR))−k2Lq−k3H(qi−qd⊗1n)(3-4)

论文中所给出的协议为 (3-4)，但个人认为可不必使用克罗内克积的形式。因为常量 q˙d\dot{q}_dq˙d 与列向量 1n1_n1n 的克罗内克积并没有特殊变化。

并且，协议 (3-4) 并不严谨，第四项中的 qiq_iqi 应为 qqq。修改版如下：
u=q˙d⊗1n−k1(Lp−diag(AR))−k2Lq−k3H(q−qd⊗1n)(3-4)\begin{aligned} u &= \dot{q}_d \otimes 1_n \\ &-k_1 (Lp - \text{diag}(AR)) \\ &-k_2 Lq \\ &-k_3 H (\red{q} - q_d \otimes 1_n) \end{aligned} \tag{3-4}u=q˙d⊗1n−k1(Lp−diag(AR))−k2Lq−k3H(q−qd⊗1n)(3-4)

当然，如果智能体并非在一维直线上运动，而是在二维平面或三维空间中，涉及到各个坐标轴的位置耦合成最终的速度时，这时候需要用到克罗内克积。

定义偏差向量：
p~=Lp−diag(AR)(3-6)\tilde{p} = L p -\text{diag} (AR) \tag{3-6}p~=Lp−diag(AR)(3-6)

展开如下：
[p~1p~2p~3p~4]=Lp−diag(A∗R)=Lp−[a12r21+a13r31+a14r41⋯⋯⋯⋯a21r12+a23r32+a24r42⋯⋯⋯⋯]\begin{aligned} \left[\begin{matrix} \tilde{p}_1 \\ \tilde{p}_2 \\ \tilde{p}_3 \\ \tilde{p}_4 \\ \end{matrix}\right]&= Lp - \text{diag}{(A*R)} \\ &= Lp - \left[\begin{matrix} a_{12}r_{21} + a_{13}r_{31} + a_{14}r_{41} & \cdots & \cdots & \cdots \\ \cdots & a_{21}r_{12} + a_{23}r_{32} + a_{24}r_{42} & \cdots & \cdots \\ \cdots \\ \cdots \\ \end{matrix}\right] \\ \end{aligned}⎣⎢⎢⎡p~1p~2p~3p~4⎦⎥⎥⎤=Lp−diag(A∗R)=Lp−⎣⎢⎢⎡a12r21+a13r31+a14r41⋯⋯⋯⋯a21r12+a23r32+a24r42⋯⋯⋯⋯⎦⎥⎥⎤

q~=qi−qd⊗1n(3-7)\tilde{q} = q_i - q_d \otimes 1_n \tag{3-7}q~=qi−qd⊗1n(3-7)

展开：
[q~1q~2q~3q~4]=[q1q2q3q4]−qd⊗1n=[q1q2q3q4]−[qdqdqdqd]\begin{aligned} \left[\begin{matrix} \tilde{q}_1 \\ \tilde{q}_2 \\ \tilde{q}_3 \\ \tilde{q}_4 \\ \end{matrix}\right]&= \left[\begin{matrix} {q}_1 \\ {q}_2 \\ {q}_3 \\ {q}_4 \\ \end{matrix}\right]-q_d \otimes 1_n \\ &= \left[\begin{matrix} {q}_1 \\ {q}_2 \\ {q}_3 \\ {q}_4 \\ \end{matrix}\right]- \left[\begin{matrix} {q}_d \\ {q}_d \\ {q}_d \\ {q}_d \\ \end{matrix}\right] \\ \end{aligned}⎣⎢⎢⎡q~1q~2q~3q~4⎦⎥⎥⎤=⎣⎢⎢⎡q1q2q3q4⎦⎥⎥⎤−qd⊗1n=⎣⎢⎢⎡q1q2q3q4⎦⎥⎥⎤−⎣⎢⎢⎡qdqdqdqd⎦⎥⎥⎤

根据定义的偏差向量，结合单个智能体的运动方程，并利用拉普拉斯矩阵的性质得
p~˙=Lq~(3-8)\dot{\tilde{p}} = L \tilde{q} \tag{3-8}p~˙=Lq~(3-8)

q~˙=−k1(p~(t−τ(t)))−k2Lq(t−τ(t))−k3H(qi−qd⊗1n)(3-9)\begin{aligned} \dot{\tilde{q}} = &-k_1 (\tilde{p}(t-\tau(t))) \\ &- k_2 L q(t-\tau(t)) \\ &-k_3 H (q_i - q_d \otimes 1_n) \end{aligned}\tag{3-9}q~˙=−k1(p~(t−τ(t)))−k2Lq(t−τ(t))−k3H(qi−qd⊗1n)(3-9)

写成矩阵的形式：
[p~˙q~˙]=[0nL0n−k3H][p~q~]+[0nL−k1⊗1n−k2L][p~(t−τ(t))q~(t−τ(t))](3-10)\begin{aligned} \left[\begin{matrix} \dot{\tilde{p}} \\ \dot{\tilde{q}} \\ \end{matrix}\right]&= \left[\begin{matrix} 0_n & L \\ 0_n & -k_3 H \\ \end{matrix}\right] \left[\begin{matrix} {\tilde{p}} \\ {\tilde{q}} \\ \end{matrix}\right]+ \left[\begin{matrix} 0_n & L \\ -k_1 \otimes 1_n & -k_2 L \\ \end{matrix}\right] \left[\begin{matrix} {\tilde{p}}(t-\tau(t)) \\ {\tilde{q}}(t-\tau(t)) \\ \end{matrix}\right] \end{aligned} \tag{3-10}[p~˙q~˙]=[0n0nL−k3H][p~q~]+[0n−k1⊗1nL−k2L][p~(t−τ(t))q~(t−τ(t))](3-10)

针对无时滞的情况，写成矩阵的形式为：
[p~˙q~˙]=[0nL−k1⊗1n−k2L−k3H][p~q~]\begin{aligned} \left[\begin{matrix} \dot{\tilde{p}} \\ \dot{\tilde{q}} \\ \end{matrix}\right]= \left[\begin{matrix} 0_n & L \\ -k_1 \otimes 1_n & -k_2 L-k_3 H \\ \end{matrix}\right] \left[\begin{matrix} {\tilde{p}} \\ {\tilde{q}} \\ \end{matrix}\right] \end{aligned}[p~˙q~˙]=[0n−k1⊗1nL−k2L−k3H][p~q~]

这里引申出来一个状态变量用来同时表示位置误差和速度误差
δ=[p~q~](3-11)\delta = \left[\begin{matrix} {\tilde{p}} \\ {\tilde{q}} \\ \end{matrix}\right] \tag{3-11}δ=[p~q~](3-11)

那么有
δ˙(t)=[p~˙q~˙]=[0nL0n−k3H][p~q~]+[0nL−k1⊗1n−k2L][p~(t−τ(t))q~(t−τ(t))]=M[p~q~]+N[p~(t−τ(t))q~(t−τ(t))]=Mδ(t)+Nδ(t−τ(t))(3-12)\begin{aligned} \dot{\delta}(t) = \left[\begin{matrix} \dot{\tilde{p}} \\ \dot{\tilde{q}} \\ \end{matrix}\right]&= \left[\begin{matrix} 0_n & L \\ 0_n & -k_3 H \\ \end{matrix}\right] \left[\begin{matrix} {\tilde{p}} \\ {\tilde{q}} \\ \end{matrix}\right]+ \left[\begin{matrix} 0_n & L \\ -k_1 \otimes 1_n & -k_2 L \\ \end{matrix}\right] \left[\begin{matrix} {\tilde{p}}(t-\tau(t)) \\ {\tilde{q}}(t-\tau(t)) \\ \end{matrix}\right] \\ &=M \left[\begin{matrix} {\tilde{p}} \\ {\tilde{q}} \\ \end{matrix}\right]+ N \left[\begin{matrix} {\tilde{p}}(t-\tau(t)) \\ {\tilde{q}}(t-\tau(t)) \\ \end{matrix}\right] \\ &=M \delta(t) + N \delta(t-\tau(t)) \end{aligned} \tag{3-12}δ˙(t)=[p~˙q~˙]=[0n0nL−k3H][p~q~]+[0n−k1⊗1nL−k2L][p~(t−τ(t))q~(t−τ(t))]=M[p~q~]+N[p~(t−τ(t))q~(t−τ(t))]=Mδ(t)+Nδ(t−τ(t))(3-12)

若是无时滞，那么为
δ˙(t)=[p~˙q~˙]=[0nL−k1⊗1n−k2L−k3H][p~q~]=MNδ(t)\begin{aligned} \dot{\delta}(t) &= \left[\begin{matrix} \dot{\tilde{p}} \\ \dot{\tilde{q}} \\ \end{matrix}\right]= \left[\begin{matrix} 0_n & L \\ -k_1 \otimes 1_n & -k_2 L-k_3 H \\ \end{matrix}\right] \left[\begin{matrix} {\tilde{p}} \\ {\tilde{q}} \\ \end{matrix}\right] \\ &=\red{M_N} \delta(t) \end{aligned} δ˙(t)=[p~˙q~˙]=[0n−k1⊗1nL−k2L−k3H][p~q~]=MNδ(t)

3.4.2 稳定性分析

选取 Lyapunov-Krasosvskii 函数
V(t)=V1+V2+V3=δT(t)Pδ(t)+∫t−τtδT(s)Qδ(s)ds+∫−m0∫t+θtδ˙T(s)Zδ˙(s)dsdθ(3-15)\begin{aligned} V(t) &= V_1+V_2+V_3\\ &= \delta^T(t) P \delta(t) \\ &+\int_{t-\tau}^{t} \delta^T(s) Q \delta(s) ~ds \\ &+\int_{-m}^{0} \int_{t+\theta}^{t} \dot{\delta}^T(s) Z \dot{\delta}(s)~ ds ~d\theta \end{aligned}\tag{3-15}V(t)=V1+V2+V3=δT(t)Pδ(t)+∫t−τtδT(s)Qδ(s) ds+∫−m0∫t+θtδ˙T(s)Zδ˙(s) ds dθ(3-15)

线形二次型求导https://blog.csdn.net/u012604810/article/details/84259965

V(x)=xTPxV(x) = x^T P xV(x)=xTPx

V˙(x)=xT(ATP+PA)x\dot{V}(x) = x^T (A^T P + P A) xV˙(x)=xT(ATP+PA)x

V˙1=2δT(t)Pδ˙(t)=2δT(t)P(Mδ(t)+Nδ(t−τ(t)))=δT(t)PMδ(t)+δT(t)MTPδ(t)+δT(t)PNδ(t−τ)+δT(t−τ)NTPδ(t)\begin{aligned} \dot{V}_1 &= 2~~ \delta^T(t) P \dot{\delta}(t) \\ &= 2 ~~\delta^T(t) P (M \delta(t) + N \delta(t-\tau(t))) \\ %&=2 ~~\delta^T(t) PM \delta(t) ~~+~~ \delta^T(t) PN \delta(t-\tau(t)) \\ %&= 2P \delta(t) \cdot M \delta(t) + 2P \delta(t) \cdot N \delta(t-\tau(t)) \\ &= \delta^T(t) PM \delta(t) + \delta^T(t) M^TP \delta(t) + \delta^T(t) PN \delta(t-\tau)+ \delta^T(t-\tau) N^TP \delta(t) \end{aligned}V˙1=2 δT(t)Pδ˙(t)=2 δT(t)P(Mδ(t)+Nδ(t−τ(t)))=δT(t)PMδ(t)+δT(t)MTPδ(t)+δT(t)PNδ(t−τ)+δT(t−τ)NTPδ(t)

定积分求导https://zhidao.baidu.com/question/940652843635700092.html

V˙2=t′⋅δT(t)Qδ(t)−(t−τ)′⋅δT(t−τ)Qδ(t−τ)=1⋅δT(t)Qδ(t)−(1−τ′)⋅δT(t−τ)Qδ(t−τ)=δT(t)Qδ(t)−(1−τ′)⋅δT(t−τ)Qδ(t−τ)≤δT(t)Qδ(t)−(1−μ)⋅δT(t−τ)Qδ(t−τ)\begin{aligned} \dot{V}_2 &= {t'} \cdot \delta^T(t) Q \delta(t) - {(t-\tau)'} \cdot \delta^T(t-\tau) Q \delta(t-\tau) \\ &= {1} \cdot \delta^T(t) Q \delta(t) - {(1-\tau')} \cdot \delta^T(t-\tau) Q \delta(t-\tau) \\ &= \delta^T(t) Q \delta(t) - {(1-\tau')} \cdot \delta^T(t-\tau) Q \delta(t-\tau) \\ &\le \delta^T(t) Q \delta(t) - {(1-\mu)} \cdot \delta^T(t-\tau) Q \delta(t-\tau) \end{aligned}V˙2=t′⋅δT(t)Qδ(t)−(t−τ)′⋅δT(t−τ)Qδ(t−τ)=1⋅δT(t)Qδ(t)−(1−τ′)⋅δT(t−τ)Qδ(t−τ)=δT(t)Qδ(t)−(1−τ′)⋅δT(t−τ)Qδ(t−τ)≤δT(t)Qδ(t)−(1−μ)⋅δT(t−τ)Qδ(t−τ)

二重积分求导https://zhidao.baidu.com/question/2142476896569356108.html

V˙3=∫−m0[∫t+θtδ˙T(s)Zδ˙(s)ds]′dθ=∫−m0[(t)′⋅δ˙T(t)Zδ˙(t)−(t+θ)′⋅δ˙T(t+θ)Zδ˙(t+θ)]dθ=∫−m0[δ˙T(t)Zδ˙(t)−(1+θ′)⋅δ˙T(t+θ)Zδ˙(t+θ)]dθ=∫−m0[δ˙T(t)Zδ˙(t)]dθ−∫−m0[(1+θ′)⋅δ˙T(t+θ)Zδ˙(t+θ)]dθ=V˙31−V˙32\begin{aligned} \dot{V}_3 &= \int_{-m}^{0} [\int_{t+\theta}^{t} \dot{\delta}^T(s) Z \dot{\delta}(s)~ ds]' ~d\theta \\ &=\int_{-m}^{0} [(t)' \cdot \dot{\delta}^T(t) Z \dot{\delta}(t) - (t+\theta)' \cdot \dot{\delta}^T(t+\theta) Z \dot{\delta}(t+\theta)] ~d\theta \\ &=\int_{-m}^{0} [\dot{\delta}^T(t) Z \dot{\delta}(t) - (1+\theta') \cdot \dot{\delta}^T(t+\theta) Z \dot{\delta}(t+\theta)] ~d\theta \\ &=\int_{-m}^{0} [\dot{\delta}^T(t) Z \dot{\delta}(t)]d\theta -\int_{-m}^{0} [(1+\theta') \cdot \dot{\delta}^T(t+\theta) Z \dot{\delta}(t+\theta)] ~d\theta \\ &= \dot{V}_{31} - \dot{V}_{32} \end{aligned}V˙3=∫−m0[∫t+θtδ˙T(s)Zδ˙(s) ds]′ dθ=∫−m0[(t)′⋅δ˙T(t)Zδ˙(t)−(t+θ)′⋅δ˙T(t+θ)Zδ˙(t+θ)] dθ=∫−m0[δ˙T(t)Zδ˙(t)−(1+θ′)⋅δ˙T(t+θ)Zδ˙(t+θ)] dθ=∫−m0[δ˙T(t)Zδ˙(t)]dθ−∫−m0[(1+θ′)⋅δ˙T(t+θ)Zδ˙(t+θ)] dθ=V˙31−V˙32

V˙31=∫−m0[δ˙T(t)Zδ˙(t)]dθ=δ˙T(t)Zδ(t)⋅(0−(−m))=mδT(t)Zδ˙(t)\begin{aligned} \dot{V}_{31} &=\int_{-m}^{0} [\dot{\delta}^T(t) Z \dot{\delta}(t)]d\theta \\ &=\dot{\delta}^T(t) Z {\delta}(t) \cdot (0-(-m)) \\ &=m {\delta}^T(t) Z \dot{\delta}(t) \end{aligned}V˙31=∫−m0[δ˙T(t)Zδ˙(t)]dθ=δ˙T(t)Zδ(t)⋅(0−(−m))=mδT(t)Zδ˙(t)

V˙32=∫−m0[(1+θ′)⋅δ˙T(t+θ)Zδ˙(t+θ)]dθ=∫t−τtδ˙T(s)Zδ˙(s)ds\begin{aligned} \dot{V}_{32} &=\int_{-m}^{0} [(1+\theta') \cdot \dot{\delta}^T(t+\theta) Z \dot{\delta}(t+\theta)] ~d\theta \\ &= \int_{t-\tau}^{t} \dot{\delta}^T(s) Z \dot{\delta}(s) ds \\ %&= \end{aligned}V˙32=∫−m0[(1+θ′)⋅δ˙T(t+θ)Zδ˙(t+θ)] dθ=∫t−τtδ˙T(s)Zδ˙(s)ds

根据引理1和引理2
V˙(t)≤V˙1(t)+V˙2(t)+V˙3(t)+mξ1T(t)Xξ(t)−∫t−τtξ1T(t)Xξ1(t)ds⏟≥0+2[δT(t)Y1+δT(t−τ)Y2][δ(t)−δ(t−τ)−∫t−τtδ˙(s)ds]⏟=0=ξ1T(t)Θξ1(t)⏟<0−∫t−τtξ2T(t)Ψξ2(t)ds⏟>0\begin{aligned} \dot{V}(t) &\le \dot{V}_1(t) + \dot{V}_2(t) + \dot{V}_3(t) \\ &+ \underbrace{m \xi_1^T(t)X \xi(t) - \int_{t-\tau}^{t} \xi_1^T(t) X \xi_1(t) ds}_{\ge0} \\ &+ \underbrace{2[\delta^T(t) Y_1 + \delta^T(t-\tau)Y_2][\delta(t)-\delta(t-\tau)-\int_{t-\tau}^{t} \dot{\delta}(s) ds]}_{=0} \\ &=\underbrace{\xi_1^T(t) \Theta \xi_1(t)}_{<0} - \underbrace{\int_{t-\tau}^{t} \xi_2^T(t) \Psi \xi_2(t) ds}_{>0} \end{aligned}V˙(t)≤V˙1(t)+V˙2(t)+V˙3(t)+≥0mξ1T(t)Xξ(t)−∫t−τtξ1T(t)Xξ1(t)ds+=02[δT(t)Y1+δT(t−τ)Y2][δ(t)−δ(t−τ)−∫t−τtδ˙(s)ds]=<0ξ1T(t)Θξ1(t)−>0∫t−τtξ2T(t)Ψξ2(t)ds

如果上述两个矩阵分别小于大于0，那么等式的右边部分整体小于0，最终 V˙(t)<0\dot{V}(t)<0V˙(t)<0 恒成立。即满足李雅普诺夫稳定性。

3.5 仿真实验与分析

3.5.1 队形保持